Integration by Parts

Integration by parts

The integration by parts applies if the integrand consists two parts (functions) in which one function is NOT the derivative of other. This rule is a special rule based on product rule for derivatives:
\( \left( uv \right)'=uv'+u'v\)
Integrate both sides and rearrange, we get
\( \int \left( uv \right)'dx=\int uv'dx+\int u'vdx\)
or \( uv=\int uv'dx+\int u'vdx\)
or \( \int uv'dx=uv-\int u'vdx\)
Integrate \( v\) as a function, so the left side remain in product form
\( \int (uv) dx=u\int vdx-\int u'\left( \int vdx \right)dx\)
To identify the first function (u) and second function (v), we Follow ILATE principle.

  1. I: Inverse trigonometric functions such as \( \sin^{-1}(x), \cos ^{-1}(x), \tan ^{-1}(x)\)
  2. L: Logarithmic functions such as \(log(x), log(x)\)
  3. A: Algebraic functions such as \(x^2, x^3\)
  4. T: Trigonometric functions such as \(\sin (x), \cos (x), \tan (x)\)
  5. E: Exponential functions such as \(e^x, 3^x\)
The ILATE principle is based on the easyness of derivative and integration.
  1. The derivative is easier when function comes from Right to Left
  2. The integral is easier when function goes from Left to Right
  3. If both function are of same type, we choose first function the one, wwhose derivative is easier (vanishes at some order), or we choose second function the one, wwhose integral is easier.
  4. Is ILATE principle compoulsary? NO, we can solve without ILATE principle as well, if applicable

Solved Examples

Evaluate the following integrals
  1. \( \int x^2 \sin x dx\)

    Solution 👉 Click Here

  2. \( \int x^2 \sin x dx\)

    Solution 👉 Click Here

  3. \( \int x^2 \sin ^{-1} xdx\)

    Solution 👉 Click Here

  4. \( \int \frac{\sqrt{x^2+1}[\log (x^2+1)-2\log x]}{x^4}dx\)
    \( \int \frac{\sqrt{x^2+1}\log (1+\frac{1}{x^2})}{x^4}dx\)
    \( \int \frac{\sqrt{x^2+1}}{x} \frac{\log (1+\frac{1}{x^2})}{x^3}dx\)
    Put \( 1+\frac{1}{x^2}=u\) then \(\frac{dx}{x^3}=-\frac{1}{2}du\)
  5. \( \tan^{-1}\sqrt{x} dx\)
    Put \(\sqrt{x}=u\)

Exercise: Evaluate the following integrals

SN Question Answer
1 \( \int x^2\cos xdx\) \( x^2\sin x-2\sin x+2x\cos x+c\)
2 \( \int x\cos xdx\) \( \cos x+x\sin x+c\)
3 \( \int x^3\sin xdx\) \( -x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+c\)
4 \( \int xe^{x^2}dx \) \( \frac{1}{2} e^{x^2}+c\)
5 \(\int \sec^2 x \csc^2 xdx \) \( \sec x \csc x-2 \cot x+c\)
6 \( \int \log xdx\) \( x \log x-x+c\)
7 \( \int x\tan^{-1} xdx\) \( \frac{1}{2}(x^2\tan^{-1} x+\tan^{-1} x-x)+c\)
8 \( \int xe^xdx \) \( (x-1)e^x+c\)
9 \( \int x^3\cos xdx\) \( x^3\sin x+3x^2\cos x-6x\sin x-6\cos x+c\)
10 \( \int{x e^x}dx\)
11 \( \int x\sin x\cos xdx \) \(\frac{x}{4}-\frac{x\cos^ 2x}{2}+\frac{\cos x\sin x}{4}+c\)
12 \( \int \sin^ 2xdx \) \( \frac{x}{2} -\frac{\sin (2x)}{4}+c\)
13 \( \int x\sin^ 2xdx\) \(\frac{x^2}{4}-\frac{\cos^ 2x}{4}-\frac{x\sin x\cos x}{2}+c\)

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